3使用ajax和php下拉

Question

國家，州和地區共有3個下拉菜單。 當我從“國家/地區”下拉列表中選擇一個國家/地區時，我所做的就是，狀態下拉列表僅顯示與該國家/地區相鄰；當我選擇“州/地區”時，地區下拉列表僅顯示與該國家/地區下拉列表相鄰的地區。

我的home.php代碼是

<?php
include "config.php";
?>
<html> 
<head> 
<script type="text/javascript">
        function showstate(str){
            if (str == "") {
                document.getElementById("state").innerHTML = "";
                return;
            }
            if (window.XMLHttpRequest) {
                // code for IE7+, Firefox, Chrome, Opera, Safari
                xmlhttp = new XMLHttpRequest();
            }
            else {// code for IE6, IE5
                xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
            }
            xmlhttp.onreadystatechange = function(){
                if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                    document.getElementById("state").innerHTML = xmlhttp.responseText;
                }
            }
            xmlhttp.open("GET", "state.php?country=" + str, true);
            xmlhttp.send();
        }
        function showcity(str2){
            if (str2 == "") {
                document.getElementById("city").innerHTML = "";
                return;
            }
            if (window.XMLHttpRequest) {
                // code for IE7+, Firefox, Chrome, Opera, Safari
                xmlhttp = new XMLHttpRequest();
            }
            else {// code for IE6, IE5
                xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
            }
            xmlhttp.onreadystatechange = function(){
                if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                    document.getElementById("city").innerHTML = xmlhttp.responseText;
                }
            }
            xmlhttp.open("GET", "district.php?state=" + str2, true);
            xmlhttp.send();
        }
 </script>
</head>
<body> 
<form > 
Choose your country : <select name="country"  onchange="showstate(this.value)"> <option>Select country  </option> 
 <?php

   $query = "SELECT DISTINCT country FROM area";
   $result = mysql_query($query);
   while($row = mysql_fetch_array($result)){
   echo"<option  value =".$row[0]."> ".$row[0]."</option>";
                                }   
 ?>
</select>
State : <select id="state" name="state" onChange="showcity(this.value)"> <option>Select state</option> </select> 
District : <select id="city" name="city"> <option>Select district </option> </select>
</form>
</body>
</html>

和state.php是

<?php
include "config.php";
echo " <option>Select state  </option> " ;
if(isset( $_GET['country']) ) 
{
  $country = $_GET['country'];
}
  $query = "SELECT DISTINCT state FROM area WHERE country = '".$country."' "; 
  $result = mysql_query($query);
    while ($row = mysql_fetch_assoc($result)) 
    {
           echo"<option  value =".$row['state']."> ".$row['state']."</option>";
    }
?>

而district.php是

<?php
include "config.php";
echo " <option>Select district  </option> " ;
if(isset( $_GET['state']) ) 
{
  $state = $_GET['state'];
}
   $query = "SELECT DISTINCT district FROM area WHERE state = '".$state."' "; 
   $result = mysql_query($query);
   while ($row = mysql_fetch_assoc($result)) 
   {

      echo"<option  value =".$row['district']."> ".$row['district']."</option>";

   }
 ?>

一切都好。 但是問題是，當我選擇任何地區時都使用所有三個下拉菜單，如果再次在同一頁面中從國家下拉菜單中重新選擇國家，則地區不會保持不變。

Answer 1

在“更改國家/地區”並在調用ur ajax調用加載狀態之前，第1次將區域組合設為空。

3使用ajax和php下拉

問題描述

1 個解決方案

解決方案1
0 2013-11-16 11:56:36

3使用ajax和php下拉

問題描述

1 個解決方案

解決方案1 0 2013-11-16 11:56:36

解決方案1
0 2013-11-16 11:56:36